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  • Language
    C#
  • License
    MIT License
  • Created over 7 years ago
  • Updated about 2 months ago

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Repository Details

Discriminated Json Subtypes Converter implementation for .NET

JsonSubTypes

JsonSubTypes is a discriminated Json sub-type Converter implementation for .NET

Build status Code Coverage Quality Gate Status NuGet NuGet CodeFactor FOSSA Status

DeserializeObject with custom type property name

[JsonConverter(typeof(JsonSubtypes), "Kind")]
public interface IAnimal
{
    string Kind { get; }
}

public class Dog : IAnimal
{
    public string Kind { get; } = "Dog";
    public string Breed { get; set; }
}

public class Cat : IAnimal {
    public string Kind { get; } = "Cat";
    public bool Declawed { get; set;}
}

The second parameter of the JsonConverter attribute is the JSON property name that will be use to retreive the type information from JSON.

var animal = JsonConvert.DeserializeObject<IAnimal>("{\"Kind\":\"Dog\",\"Breed\":\"Jack Russell Terrier\"}");
Assert.AreEqual("Jack Russell Terrier", (animal as Dog)?.Breed);

N.B.: This only works for types in the same assembly as the base type/interface and either in the same namespace or with a fully qualified type name.

DeserializeObject with custom type mapping

[JsonConverter(typeof(JsonSubtypes), "Sound")]
[JsonSubtypes.KnownSubType(typeof(Dog), "Bark")]
[JsonSubtypes.KnownSubType(typeof(Cat), "Meow")]
public class Animal
{
    public virtual string Sound { get; }
    public string Color { get; set; }
}

public class Dog : Animal
{
    public override string Sound { get; } = "Bark";
    public string Breed { get; set; }
}

public class Cat : Animal
{
    public override string Sound { get; } = "Meow";
    public bool Declawed { get; set; }
}
var animal = JsonConvert.DeserializeObject<IAnimal>("{\"Sound\":\"Bark\",\"Breed\":\"Jack Russell Terrier\"}");
Assert.AreEqual("Jack Russell Terrier", (animal as Dog)?.Breed);

N.B.: Also works with other kind of value than string, i.e.: enums, int, ...

SerializeObject and DeserializeObject with custom type property only present in JSON

This mode of operation only works when JsonSubTypes is explicitely registered in JSON.NET's serializer settings, and not through the [JsonConverter] attribute.

public abstract class Animal
{
    public int Age { get; set; }
}

public class Dog : Animal
{
    public bool CanBark { get; set; } = true;
}

public class Cat : Animal
{
    public int Lives { get; set; } = 7;
}

public enum AnimalType
{
    Dog = 1,
    Cat = 2
}

Registration:

var settings = new JsonSerializerSettings();
settings.Converters.Add(JsonSubtypesConverterBuilder
    .Of(typeof(Animal), "Type") // type property is only defined here
    .RegisterSubtype(typeof(Cat), AnimalType.Cat)
    .RegisterSubtype(typeof(Dog), AnimalType.Dog)
    .SerializeDiscriminatorProperty() // ask to serialize the type property
    .Build());

or using syntax with generics:

var settings = new JsonSerializerSettings();
settings.Converters.Add(JsonSubtypesConverterBuilder
    .Of<Animal>("Type") // type property is only defined here
    .RegisterSubtype<Cat>(AnimalType.Cat)
    .RegisterSubtype<Dog>(AnimalType.Dog)
    .SerializeDiscriminatorProperty() // ask to serialize the type property
    .Build());

De-/Serialization:

var cat = new Cat { Age = 11, Lives = 6 }

var json = JsonConvert.SerializeObject(cat, settings);

Assert.Equal("{\"Lives\":6,\"Age\":11,\"Type\":2}", json);

var result = JsonConvert.DeserializeObject<Animal>(json, settings);

Assert.Equal(typeof(Cat), result.GetType());
Assert.Equal(11, result.Age);
Assert.Equal(6, (result as Cat)?.Lives);

DeserializeObject mapping by property presence

[JsonConverter(typeof(JsonSubtypes))]
[JsonSubtypes.KnownSubTypeWithProperty(typeof(Employee), "JobTitle")]
[JsonSubtypes.KnownSubTypeWithProperty(typeof(Artist), "Skill")]
public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

public class Employee : Person
{
    public string Department { get; set; }
    public string JobTitle { get; set; }
}

public class Artist : Person
{
    public string Skill { get; set; }
}

or using syntax with generics:

string json = "[{\"Department\":\"Department1\",\"JobTitle\":\"JobTitle1\",\"FirstName\":\"FirstName1\",\"LastName\":\"LastName1\"}," +
                "{\"Department\":\"Department1\",\"JobTitle\":\"JobTitle1\",\"FirstName\":\"FirstName1\",\"LastName\":\"LastName1\"}," +
                "{\"Skill\":\"Painter\",\"FirstName\":\"FirstName1\",\"LastName\":\"LastName1\"}]";


var persons = JsonConvert.DeserializeObject<IReadOnlyCollection<Person>>(json);
Assert.AreEqual("Painter", (persons.Last() as Artist)?.Skill);

Registration:

settings.Converters.Add(JsonSubtypesWithPropertyConverterBuilder
    .Of(typeof(Person))
    .RegisterSubtypeWithProperty(typeof(Employee), "JobTitle")
    .RegisterSubtypeWithProperty(typeof(Artist), "Skill")
    .Build());

or

settings.Converters.Add(JsonSubtypesWithPropertyConverterBuilder
    .Of<Person>()
    .RegisterSubtypeWithProperty<Employee>("JobTitle")
    .RegisterSubtypeWithProperty<Artist>("Skill")
    .Build());

A default class other than the base type can be defined

[JsonConverter(typeof(JsonSubtypes))]
[JsonSubtypes.KnownSubType(typeof(ConstantExpression), "Constant")]
[JsonSubtypes.FallBackSubType(typeof(UnknownExpression))]
public interface IExpression
{
    string Type { get; }
}

Or with code configuration:

settings.Converters.Add(JsonSubtypesConverterBuilder
    .Of(typeof(IExpression), "Type")
    .SetFallbackSubtype(typeof(UnknownExpression))
    .RegisterSubtype(typeof(ConstantExpression), "Constant")
    .Build());
settings.Converters.Add(JsonSubtypesWithPropertyConverterBuilder
    .Of(typeof(IExpression))
    .SetFallbackSubtype(typeof(UnknownExpression))
    .RegisterSubtype(typeof(ConstantExpression), "Value")
    .Build());

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License

FOSSA Status